Question

PLEASE HELP!! In triangle ABC, B= 120 degrees, a=10, c=18, find C

Answer 1

Add all of them then subtract from 180

Answer 2

Check the picture below.  Notice that we can always solve for the cosine and simply use the sides alone to get an angle.

$\bf cos^{-1}\left(\cfrac{10^2+(\sqrt{604})^2-18^2}{2(10)(\sqrt{604})} \right)=\measuredangle C \\\\\\ cos^{-1}\left(\cfrac{100+604-324}{20\sqrt{604}} \right)=\measuredangle C \\\\\\ cos^{-1}(0.77309903583)\approx \measuredangle C\implies 39.366998916677^o\approx \measuredangle C$